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# mathproblems131to140 ( ** )

### Mathematical problems — Problems 131 to 140

The ten problems are presented first, followed by the solutions to the ten problems.

Problem 131:

From a point on the ground 33 feet away from the base of a building, the distance from that point to the top of a window in the building is 65 feet, and the distance from the same point to the bottom of the window is 55 feet.  How tall is the window?

Problem 132:

One high-speed train leaves the train station at noon heading north at 80 miles per hour.  Another high speed train leaves the station at 2 pm heading west.  At 3 pm the two trains are 250 miles apart.  What is the speed of the west-bound train?

Problem 133:

Two planes leave the same airport at the same time.  One plane travels east at 190 mph but faces a headwind blowing at 40 mph.  The other plane travels north and has a tailwind of 70 mph.  After one hour of travel, the planes are 250 miles apart.  What is the travel speed of the northbound plane?

Problem 134:

A rancher owns horses and roosters.  He owns 30 animals, and the total number of legs of his animals is 74.  How many horses and how many roosters does he own?

Problem 135:

A flower garden is to contain 24 plants, made up of roses, carnations, and pansies.  The gardener wants twice as many carnations as roses and doesn’t want to spend more than \$8.20 for the plants.  If rose plants cost 50 cents each, carnations 25 cents each, and pansies 40 cents each, then what combination of flower plants should be used to match the maximum price?

Problem 136:

A service group is selling candy bars and bags of almonds to raise money.  They are selling the bars for \$1 and the bags for \$1.25.  The group paid 40 cents for the bars and 50 cents for the bags.  Their total revenue for the sale of 1,350 items was \$1,500.00.  How many of each item did they sell, and what was their net profit?

Problem 137:

Michael is 71 years old, and he runs 4 half-marathons per year.  He has already completed 50 half-marathons.   How old will he be when the number of races run will equal his age?

Problem 138:

Ten times the sum of two consecutive integers is 28 more than the product of the two numbers.  What are the numbers?

Problem 139:

Mark can mow a certain lawn in 30 minutes, and it takes Steven 60 minutes to mow the same lawn.  Working together, how long would it take for them to mow the lawn?

Problem 140:

There are 115 marbles in a bag.  The number of black marbles is 4 more than twice the number of red marbles.  How many marbles of each color are in the bag?

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Solution to Problem 131:

• We are dealing with two right triangles, with the adjacent side of both triangles = 33 feet.
• The hypotenuse of the smaller triangle = 55 feet and the hypotenuse of the larger triangle = 65 feet.
• Also, the length of the opposite side of the smaller triangle plus the height of the window = length of the opposite side of the larger triangle.
• The opposite side of the smaller window = √(55² – 33²) = 44 feet.  The opposite side of the larger triangle = √(65² – 33²) = 56 feet.
• Thus, the window must be 12 feet in height (56 – 44).

Solution to Problem 132:

• The first train has traveled north for 3 hours at 80 miles per hour, so it has traveled 240 miles.  This is one side of a right triangle, with 250 miles being the hypotenuse.
• The other train has traveled one hour at x mph, so it has traveled x miles.  Thus,  x² + 240² = 250².
• Thus, x = √ (62,500 – 57,600)  = √ (4,900)  = 70.  Thus, the westbound train is traveling at 70 miles per hour.

Solution to Problem 133:

• We are dealing with a right triangle, so Pythagorean’s theorem comes into play.
• We know that (190 – 40)² + (x + 70)² =  250², with x = speed of the northbound plane.
• Thus,  22,500 + x² + 70x + 4900 = 62,500, which simplifies to  x² + 70x -31,500 = 0.
• This factors into (x + 270)(x – 130) = 0.  Thus, the speed of the plane is 130 mph.

Solution to Problem 134:

• Let h = the number of horses and r = the number of roosters.  We know horses have 4 legs and roosters have 2 legs.
• Thus, 4h + 2r = 74, and we know that h + r = 30.   Solving for h yields  h = 30 – r.
• Substituting gives 4(30 – r) + 2r = 74.  Thus, 120 – 4r + 2r = 74.  This yields  2r = 46, and r = 23.
• Thus the number of roosters equals 23 and the number of horses equals 7.

Solution to Problem 135:

• Let r = rose plants, c = carnation plants, and p = pansie plants.
• We know that 2r = c and also that r + c + p = 24.  Substitution leads to  3r + p = 24, or p = 24 – 3r
• The pricing requirements lead to  0.5r + 0.25c + 0.4p = 8.20
• Substitution yields  0.5r + 0.25(2r) + 0.4(24 – 3r) = 8.20.
• Simplification yields  0.5 r + 0.5r + 9.6 – 1.2r = 8.20, and thus 0.2r = 1.4, and r = 7
• If r = 7, then c = 14, and p = 3.  Thus, 7 rose plants, 14 carnation plants, and 3 pansie plants.

Solution to Problem 136:

• Let b = bars and a = almond bags.  Thus, b + a = 1350, or b = 1350 – a.
• We know that 1(b) + 1.25(a) = 1500.  Substitution yields  1(1350 – a) + 1.25(a) = 1500.
• Simplification yields 0.25a = 150.  Thus a = 600 and b = 750.
• To calculate the profit, we know that 0.4(750) + 0.5(600) = 300 + 300 = \$600.
• Then, \$1,500 – \$600 = \$900, which is the net profit.

Solution to Problem 137:

• Let x = the number of years that will elapse before reaching the goal.
• Thus,  71 + x = 50 + 4x.  Simplification yields  3x = 21.
• Thus, in 7 years Michael will be 78 years old and will have run 78 half-marathons.

Solution to Problem 138:

• Let a = the smaller integer and let b = the larger integer.
• We know that  10(a + b) – 28 = a · b.  We also know that  b = a + 1.
• Substitution yields  10(a + (a + 1)) – 28 = a (a + 1).
• Simplification yields 10(2a + 1) = a² + a + 28, and then 20a + 10 = a² + a + 28.
• Simplification yields  a² – 19a + 18 =  0.  Factoring gives (a – 18)(a – 1) = 0.
• Thus, there are two solutions.  The numbers can be 1 and 2 or 18 and 19.

Solution to Problem 139:

• An equation can be written thusly:  (1 lawn / 30 minutes) + (1 lawn / 60 minutes) = (1 lawn / x minutes)
• Thus, 1/30 + 1/60 = 1/x.   Then,  30 • 60 (1/30 + 1/60) = (1/x) 30 • 60. Simplification yields  60 + 30 = 1800 / x
• And  x ( 90 ) = 1800.  Thus x = 20.  The two workers could mow the lawn in 20 minutes.

Solution to Problem 140:

• Let r = the number of red marbles and b = the number of black marbles.
• We know that b + r = 115, and also that 2r + 4 = b.
• Substituting yields  (2r + 4) + r = 115.  Solving for r yields 3r = 111.
• Thus, r = 37 and b = 78.